Given a line $y=\theta(1-x)$ where $0<x<1$, $0<y<1$ and $0<\theta<1$, I have a collection of curves
$$
y^K=1-(1-x)^K
$$
parametrized by an integer $K>2$. For any $K$, the related curve intersects the line at only one point. For this intersection point let
$$
A=y^K
$$
On the other hand I have
\begin{equation}
\prod_{i=1}^K {y_i}=1-\prod_{i=1}^K(1-x_i)
\end{equation}
where $x_i$ and $y_i$ are some points on $y=\theta(1-x)$ and let
$$
B=\prod_{i=1}^K {y_i}
$$
Is it possible that $B<A$?

1 Answer
1

Replace $1-x$ by $x$. Then you have the line $\ \ell\colon\ y=\theta x$ and for a given $K>2$ the curve
$$\gamma: \quad x^K+y^K=1\ ,$$
where everything is restricted to the window $[0,1]^2$. The curve $\gamma$ intersects $\ell$ in a point $p:=(u,v)$. Obviously $v=\theta u$, and as $p\in\gamma$ we get the equation
$$u^K(1+\theta^K)=1\ .$$
This implies
$$A:=v^K=\theta^K u^K={\theta^K\over 1+\theta^K}\ .$$
On the other hand you are given $K$ points $(x_i,y_i)\in\ell$, chosen such that $$\prod_{i=1}^K x_i + \prod_{i=1}^K y_i=1\ .$$
As $y_i=\theta x_i$ for each $i$ we have $(\theta^{-K}+1)\prod_{i=1}^K y_i=1$ or
$$B:=\prod_{i=1}^K y_i={1\over \theta^{-K}+1}={\theta^K\over 1+\theta^K}=A\ .$$